If I have a finite theory with a signature with finitely many constant, function, and relation symbols, then I can sometimes (possibly always) formulate an equivalent theory associated with a signature with a single relation symbol.
I have an explicit 4-place relation symbol that can express the theory of commutative rings $Q(x, y, z, w) \leftrightarrow w=(x+y)\cdot z $.
I'm wondering whether this is the best we can do, or whether there's a three-place relation that can capture the theory of rings.
I'm interested in whether rings can be expressed using a three-place relation symbol.
For example, the theory of groups has the following signature $\{e, {}^{-1}, \cdot\}$, but all of these can be expressed in terms of a single ternary relation.
$$ R(x, y, z) \;\; \text{if and only if} \;\; x \cdot y = z $$
The identity element in the sole element in the domain satisfying the following
$$ \forall x \mathop R(e, x, x) $$
The inverse of $x$ is $y$ if and only if the following holds.
$$R(x, y, e)$$
For rings, I am pretty sure the best that I can do is a four-place relation
$$ Q(x, y, z, w) \;\; \text{if and only if} \;\; (x+y)*z = w $$
$0$ is the multiplicative annihilator. $$\forall x \forall y Q(x, y, k, k) \;\text{holds}\;\; \text{if and only if}\;\; k = 0 $$.
$1$ is the multiplicative identity.
$$ \forall x Q(x, 0, k, x) \; \text{holds} \;\; \text{if and only if} \;\; k = 1 $$
I can define the sum of $x$ and $y$.
$$ Q(x, y, 1, z) \;\; \text{if and only if $z$ is the sum of $x$ and $y$} $$
I can define the product of $x$ and $y$.
$$ Q(x, 0, y, z) \;\; \text{if and only if $z$ is the product of $x$ and $y$} $$
Best Answer
The operations of any ring can be defined (in first-order logic) from the relation $$x\cdot y=z\vee x+y=z.$$ Let me think of this relation as defining a binary "quasi-operation" $x*y$ that can take two different values, either $xy$ or $x+y$. I will write $x*y=z$ to mean that $z$ is one of the values of $x*y$, though of course we must be careful with this notation since this use of "$=$" is not transitive. If our ring has one element, then our task is trivial, so we may assume it has more than one element (and in particular it satisfies $0\neq 1$).
First note that we can define $0$ as the unique $a$ such that $a$ is the only value of $a*a$. Once we have $0$, we can define $1$ as the unique $a\neq 0$ such that $a*y=y$ for all $y$. We can then define $x+1$ as the value of $x*1$ that is different from $x$. Then $x-1$ is the unique $a$ such that $a+1=x$, and in particular we can define $-1=0-1$. Then we can define $-x$ as the value of $(-1)*x$ that is different from $x-1$ (or the unique value of $(-1)*x$ if there is only one). Also, we can define the condition $xy=1$ to mean that either $1$ is the only value of $x*y$ or that $x*y=1$ and $y\neq 1+(-x)$.
Finally, we can define $x+y$ as follows:
The only way the second case could fail to uniquely describe $x+y$ is if both $xy(-y)=x$ and $xy(-y+1)=x+1$. But these two equations together give $xy=1$, and so the first case handles that.
And of course, once we have $x+y$, we can define $xy$ as the value of $x*y$ other than $x+y$ (or the unique value if there is only one).